Solutions Exercise Question & Answers(Q2.1Q2.22)
 By srinivaseducation
 On 05/04/2013
 Comments (0)
 In CHEMISTRY
SOLUTIONS QUESTION AND ANSWERS (Q2.1Q2.22)
EXERCISE QUESTION ANS ANSWERS
Page : 59
Q2.1: Define the term solution. How many types of solutions are formed? Write briefly about each type with an example.
Ans: Homogeneous mixtures of two or more than two components are known as solutions.
There are three types of solutions.
(i) Gaseous solution:
The solution in which the solvent is a gas is called a gaseous solution. In these solutions, the solute may be liquid, solid, or gas. For example, a mixture of oxygen and nitrogen gas is a gaseous solution.
(ii) Liquid solution:
The solution in which the solvent is a liquid is known as a liquid solution. The solute in these solutions may be gas, liquid, or solid.
For example, a solution of ethanol in water is a liquid solution.
(iii) Solid solution:
The solution in which the solvent is a solid is known as a solid solution. The solute may be gas, liquid or solid. For example, a solution of copper in gold is a solid solution.
Q2.2: Give an example of solid solution in which the solute is a gas.
Ans: In case a solid solution is formed between two substances (one having very large particles and the other having very small particles), an interstitial solid solution will be formed. For example, a solution of hydrogen in palladium is a solid solution in which the solute is a gas.
Q2.3: Define the following terms:
(i) Mole fraction
(ii) Molality
(iii) Molarity
(iv) Mass percentage.
Ans: (i) Mole fraction:
The mole fraction of a component in a mixture is defined as the ratio of the number of moles of the component to the total number of moles of all the components in the mixture.
i.e.,
Mole fraction of a component
Mole fraction is denoted by ‘x’.
If in a binary solution, the number of moles of the solute and the solvent are n_{A} and n_{B} respectively, then the mole fraction of the solute in the solution is given by,
Similarly, the mole fraction of the solvent in the solution is given as:
(ii) Molality
Molality (m) is defined as the number of moles of the solute per kilogram of the solvent. It is expressed as:
Molality (m)
(iii) Molarity
Molarity (M) is defined as the number of moles of the solute dissolved in one Litre of the solution.
It is expressed as:
Molarity (M)
(iv) Mass percentage:
The mass percentage of a component of a solution is defined as the mass of the solute in grams present in 100 g of the solution. It is expressed as:
Mass % of a component
Q2.4; Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in aqueous solution. What should be the molarity of such a sample of the acid if the density of the solution is 1.504 g mL^{−1}?
Ans: Concentrated nitric acid used in laboratory work is 68% nitric acid by mass in an aqueous solution. This means that 68 g of nitric acid is dissolved in 100 g of the solution.
Molar mass of nitric acid (HNO_{3}) = 1 × 1 + 1 × 14 + 3 × 16 = 63 g mol^{−1}
Then, number of moles of HNO_{3 }
Given,
Density of solution = 1.504 g mL^{−1}
Volume of 100 g solution =
Molarity of solution
Q2.5: A solution of glucose in water is labelled as 10% w/w, what would be the molality and mole fraction of each component in the solution? If the density of solution is 1.2 g mL^{−1}, then what shall be the molarity of the solution?
Ans: 10% w/w solution of glucose in water means that 10 g of glucose in present in 100 g of the solution i.e., 10 g of glucose is present in (100 − 10) g = 90 g of water.
Molar mass of glucose (C_{6}H_{12}O_{6}) = 6 × 12 + 12 × 1 + 6 × 16 = 180 g mol^{−1}
Then, number of moles of glucose
= 0.056 mol
Molality of solution = 0.62 m
Number of moles of water
= 5 mol
Mole fraction of glucose
And, mole fraction of water
= 1 − 0.011
= 0.989
If the density of the solution is 1.2 g mL^{−1}, then the volume of the 100 g solution can be given as:
Molarity of the solution
= 0.67 M
Q2.6: How many mL of 0.1 M HCl are required to react completely with 1 g mixture of Na_{2}CO_{3} and NaHCO_{3} containing equimolar amounts of both?
Ans: Let the amount of Na_{2}CO_{3} in the mixture be x g.
Then, the amount of NaHCO_{3} in the mixture is (1 − x) g.
Molar mass of Na_{2}CO_{3} = 2 × 23 + 1 × 12 + 3 × 16
= 106 g mol^{−1}
Number of moles Na_{2}CO_{3}
Molar mass of NaHCO_{3} = 1 × 23 + 1 × 1 × 12 + 3 × 16
= 84 g mol^{−1}
Number of moles of NaHCO_{3}
According to the question,
⇒ 84x = 106 − 106x
⇒ 190x = 106
⇒ x = 0.5579
Therefore, number of moles of Na_{2}CO_{3}
= 0.0053 mol
And, number of moles of NaHCO_{3}
= 0.0053 mol
HCl reacts with Na_{2}CO_{3} and NaHCO_{3} according to the following equation.
1 mol of Na_{2}CO_{3} reacts with 2 mol of HCl.
Therefore, 0.0053 mol of Na_{2}CO_{3} reacts with 2 × 0.0053 mol = 0.0106 mol.
Similarly, 1 mol of NaHCO_{3} reacts with 1 mol of HCl.
Therefore, 0.0053 mol of NaHCO_{3} reacts with 0.0053 mol of HCl.
Total moles of HCl required = (0.0106 + 0.0053) mol
= 0.0159 mol
In 0.1 M of HCl,
0.1 mol of HCl is preset in 1000 mL of the solution.
Therefore, 0.0159 mol of HCl is present in
= 159 mL of the solution
Hence, 159 mL of 0.1 M of HCl is required to react completely with 1 g mixture of Na_{2}CO_{3} and NaHCO_{3,} containing equimolar amounts of both.
Q2.7: A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass. Calculate the mass percentage of the resulting solution.
Ans: Total amount of solute present in the mixture is given by,
= 75 + 160
= 235 g
Total amount of solution = 300 + 400 = 700 g
Therefore, mass percentage (w/w) of the solute in the resulting solution,
= 33.57%
And, mass percentage (w/w) of the solvent in the resulting solution,
= (100 − 33.57)%
= 66.43%
Q2.8: An antifreeze solution is prepared from 222.6 g of ethylene glycol (C_{2}H_{6}O_{2}) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL^{−1}, then what shall be the molarity of the solution?
Ans: Molar mass of ethylene glycol= 2 × 12 + 6 × 1 + 2 ×16
= 62 gmol^{−1}
Number of moles of ethylene glycol
= 3.59 mol
Therefore, molality of the solution
= 17.95 m
Total mass of the solution = (222.6 + 200) g
= 422.6 g
Given,
Density of the solution = 1.072 g mL^{−1}
Volume of the solution
= 394.22 mL
= 0.3942 × 10^{−3} L
Molarity of the solution
= 9.11 M
Q2.9: A sample of drinking water was found to be severely contaminated with chloroform (CHCl_{3}) supposed to be a carcinogen. The level of contamination was 15 ppm (by mass):
(i) express this in percent by mass
(ii) determine the molality of chloroform in the water sample.
Ans; (i) 15 ppm (by mass) means 15 parts per million (10^{6}) of the solution.
Therefore, percent by mass
= 1.5 × 10^{−5 }%
(ii) Molar mass of chloroform (CHCl_{3}) = 1 × 12 + 1 × 1 + 3 × 35.5
= 119.5 g mol^{−1}
Now, according to the question,
15 g of chloroform is present in 10^{6} g of the solution.
i.e., 15 g of chloroform is present in (10^{6} − 15) ≈ 10^{6} g of water.
Molality of the solution
= 1.26 × 10^{−4} m
Q2.10; What role does the molecular interaction play in a solution of alcohol and water?
Ans: In pure alcohol and water, the molecules are held tightly by a strong hydrogen bonding. The interaction between the molecules of alcohol and water is weaker than alcohol−alcohol and water−water interactions. As a result, when alcohol and water are mixed, the intermolecular interactions become weaker and the molecules can easily escape. This increases the vapour pressure of the solution, which in turn lowers the boiling point of the resulting solution.
Q2.11: Why do gases always tend to be less soluble in liquids as the temperature is raised?
Ans: Solubility of gases in liquids decreases with an increase in temperature. This is because dissolution of gases in liquids is an exothermic process.
Therefore, when the temperature is increased, heat is supplied and the equilibrium shifts backwards, thereby decreasing the solubility of gases.
Q2.12: State Henry’s law and mention some important applications?
Ans: Henry’s law states that partial pressure of a gas in the vapour phase is proportional to the mole fraction of the gas in the solution. If p is the partial pressure of the gas in the vapour phase and x is the mole fraction of the gas, then Henry’s law can be expressed as:
p = K_{H }x
Where,
K_{H} is Henry’s law constant
Some important applications of Henry’s law are mentioned below.
(i) Bottles are sealed under high pressure to increase the solubility of CO_{2} in soft drinks and soda water.
(ii) Henry’s law states that the solubility of gases increases with an increase in pressure. Therefore, when a scuba diver dives deep into the sea, the increased sea pressure causes the nitrogen present in air to dissolve in his blood in great amounts. As a result, when he comes back to the surface, the solubility of nitrogen again decreases and the dissolved gas is released, leading to the formation of nitrogen bubbles in the blood. This results in the blockage of capillaries and leads to a medical condition known as ‘bends’ or ‘decompression sickness’.
Hence, the oxygen tanks used by scuba divers are filled with air and diluted with helium to avoid bends.
(iii) The concentration of oxygen is low in the blood and tissues of people living at high altitudes such as climbers. This is because at high altitudes, partial pressure of oxygen is less than that at ground level. Lowblood oxygen causes climbers to become weak and disables them from thinking clearly. These are symptoms of anoxia.
Q2.13: The partial pressure of ethane over a solution containing 6.56 × 10^{−3} g of ethane is 1 bar. If the solution contains 5.00 × 10^{−2} g of ethane, then what shall be the partial pressure of the gas?
Ans: Molar mass of ethane (C_{2}H_{6}) = 2 × 12 + 6 × 1
= 30 g mol^{−1}
Number of moles present in 6.56 × 10^{−2} g of ethane
= 2.187 × 10^{−3} mol
Let the number of moles of the solvent be x.
According to Henry’s law,
p = K_{H}x
(Since x >> 2.187 × 10^{−3})
Number of moles present in 5.00 × 10^{−2} g of ethane
= 1.67 × 10^{−3} mol
According to Henry’s law,
p = K_{H}x
(Since, x >> 1.67 × 10^{−3})
= 0.764 bar
Hence, partial pressure of the gas shall be 0.764 bar.
Q2.14: What is meant by positive and negative deviations from Raoult's law and how is the sign of Δ_{sol}H related to positive and negative deviations from Raoult's law?
Ans: According to Raoult’s law, the partial vapour pressure of each volatile component in any solution is directly proportional to its mole fraction. The solutions which obey Raoult’s law over the entire range of concentration are known as ideal solutions. The solutions that do not obey Raoult’s law (nonideal solutions) have vapour pressures either higher or lower than that predicted by Raoult’s law. If the vapour pressure is higher, then the solution is said to exhibit positive deviation, and if it is lower, then the solution is said to exhibit negative deviation from Raoult’s law.
Vapour pressure of a twocomponent solution showing positive deviation from Raoult’s law
Vapour pressure of a twocomponent solution showing negative deviation from Raoult’s law
In the case of an ideal solution, the enthalpy of the mixing of the pure components for forming the solution is zero.
Δ_{sol}H = 0
In the case of solutions showing positive deviations, absorption of heat takes place.
∴Δ_{sol}H = Positive
In the case of solutions showing negative deviations, evolution of heat takes place.
∴Δ_{sol}H = Negative
Q2.15: An aqueous solution of 2% nonvolatile solute exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute?
Ans: Here,
Vapour pressure of the solution at normal boiling point (p_{1}) = 1.004 bar
Vapour pressure of pure water at normal boiling point
Mass of solute, (w_{2}) = 2 g
Mass of solvent (water), (w_{1}) = 98 g
Molar mass of solvent (water), (M_{1}) = 18 g mol^{−1}
According to Raoult’s law,
= 41.35 g mol^{−1}
Hence, the molar mass of the solute is 41.35 g mol^{−1}.
Q2.16: Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour pressure of a mixture of 26.0 g of heptane and 35 g of octane?
Ans: Vapour pressure of heptane
Vapour pressure of octane = 46.8 kPa
We know that,
Molar mass of heptane (C_{7}H_{16}) = 7 × 12 + 16 × 1
= 100 g mol^{−1}
Number of moles of heptane
= 0.26 mol
Molar mass of octane (C_{8}H_{18}) = 8 × 12 + 18 × 1
= 114 g mol^{−1}
Number of moles of octane
= 0.31 mol
Mole fraction of heptane,
= 0.456
And, mole fraction of octane, x_{2} = 1 − 0.456
= 0.544
Now, partial pressure of heptane,
= 0.456 × 105.2
= 47.97 kPa
Partial pressure of octane,
= 0.544 × 46.8
= 25.46 kPa
Hence, vapour pressure of solution, p_{total }= p_{1} + p_{2}
= 47.97 + 25.46
= 73.43 kPa
Q2.17: The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal solution of a nonvolatile solute in it.
Ans: 1 molal solution means 1 mol of the solute is present in 1000 g of the solvent (water).
Molar mass of water = 18 g mol^{−1}
Number of moles present in 1000 g of water
= 55.56 mol
Therefore, mole fraction of the solute in the solution is
.
It is given that,
Vapour pressure of water, = 12.3 kPa
Applying the relation,
⇒ 12.3 − p_{1} = 0.2177
⇒ p_{1} = 12.0823
= 12.08 kPa (approximately)
Hence, the vapour pressure of the solution is 12.08 kPa.
Q2.18: Calculate the mass of a nonvolatile solute (molar mass 40 g mol^{−1}) which should be dissolved in 114 g octane to reduce its vapour pressure to 80%.
Ans; Let the vapour pressure of pure octane be
Then, the vapour pressure of the octane after dissolving the nonvolatile solute is
Molar mass of solute, M_{2} = 40 g mol^{−1}
Mass of octane, w_{1} = 114 g
Molar mass of octane, (C_{8}H_{18}), M_{1} = 8 × 12 + 18 × 1
= 114 g mol^{−1}
Applying the relation,
Hence, the required mass of the solute is 8 g.
Q2.19: A solution containing 30 g of nonvolatile solute exactly in 90 g of water has a
vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to
the solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:

molar mass of the solute

vapour pressure of water at 298 K.
Ans: (i) Let, the molar mass of the solute be M g mol^{−1}
Now, the no. of moles of solvent (water),
And, the no. of moles of solute,
Applying the relation:
After the addition of 18 g of water:
Again, applying the relation:
Dividing equation (i) by (ii), we have:
Therefore, the molar mass of the solute is 23 g mol^{−1}.
(ii) Putting the value of ‘M’ in equation (i), we have:
Hence, the vapour pressure of water at 298 K is 3.53 kPa.
Q2.20: A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of 5% glucose in water if freezing point of pure water is 273.15 K.
Ans: Here, ΔT_{f} = (273.15 − 271) K
= 2.15 K
Molar mass of sugar (C_{12}H_{22}O_{11}) = 12 × 12 + 22 × 1 + 11 × 16
= 342 g mol^{−1}
5% solution (by mass) of cane sugar in water means 5 g of cane sugar is present in (100 − 5)g = 95 g of water.
Now, number of moles of cane sugar
= 0.0146 mol
Therefore, molality of the solution,
= 0.1537 mol kg^{−1}
Applying the relation,
ΔT_{f} = K_{f} × m
= 13.99 K kg mol^{−1}
Molar of glucose (C_{6}H_{12}O_{6}) = 6 × 12 + 12 × 1 + 6 × 16
= 180 g mol^{−1}
5% glucose in water means 5 g of glucose is present in (100 − 5) g = 95 g of water.
Number of moles of glucose
= 0.0278 mol
Therefore, molality of the solution,
= 0.2926 mol kg^{−1}
Applying the relation,
ΔT_{f} = K_{f} × m
= 13.99 K kg mol^{−1} × 0.2926 mol kg^{−1}
= 4.09 K (approximately)
Hence, the freezing point of 5% glucose solution is (273.15 − 4.09) K= 269.06 K.
Q2.21; Two elements A and B form compounds having formula AB_{2} and AB_{4}. When dissolved in 20 g of benzene (C_{6}H_{6}), 1 g of AB_{2} lowers the freezing point by 2.3 Kwhereas 1.0 g of AB_{4} lowers it by 1.3 K. The molar depression constant for benzene is 5.1 Kkg mol^{−1}. Calculate atomic masses of A and B.
Ans: We know that,
Then,
= 110.87 g mol^{−1}
= 196.15 g mol^{−1}
Now, we have the molar masses of AB_{2} and AB_{4} as 110.87 g mol^{−1} and 196.15 g mol^{−1} respectively.
Let the atomic masses of A and B be x and y respectively.
Now, we can write:
Subtracting equation (i) from (ii), we have
2y = 85.28
⇒ y = 42.64
Putting the value of ‘y’ in equation (1), we have
x + 2 × 42.64 = 110.87
⇒ x = 25.59
Hence, the atomic masses of A and B are 25.59 u and 42.64 u respectively.
Q2.22: At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
Ans; Here,
T = 300 K
π = 1.52 bar
R = 0.083 bar L K^{−1} mol^{−1}
Applying the relation,
π = CRT
= 0.061 mol
Since the volume of the solution is 1 L, the concentration of the solution would be 0.061 M.